编写程序，完成手机计算器中表达式的求解（难度分为四级）基本要求：无括号、整数用c写

在C语言中，编写一个基础的手机计算器功能，用于解决无括号的整数加减乘除表达式，可以采用递归解析法，也就是栈的应用。这里我们简化处理只考虑四则运算，忽略优先级规则。下面是一个基础版本的示例：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node {
    int value;
    char op;
    struct Node* left;
    struct Node* right;
} Node;

Node* createNumber(int val) {
    Node* node = (Node*)malloc(sizeof(Node));
    node->value = val;
    node->left = NULL;
    node->right = NULL;
    node->op = '\0';
    return node;
}

Node* createOperator(char op, Node* left, Node* right) {
    Node* node = (Node*)malloc(sizeof(Node));
    node->value = 0;
    node->op = op;
    node->left = left;
    node->right = right;
    return node;
}

int applyOp(int a, int b, char op) {
    switch (op) {
        case '+': return a + b;
        case '-': return a - b;
        case '*': return a * b;
        case '/': return a / b; // 检查除数是否为0
        default: return 0; // 错误操作符
    }
}

int evaluateExpression(char* expression) {
    Node* root = createNumber(strtol(expression, NULL, 10));
    char* token = strtok(expression, " ");
    while(token != NULL) {
        if(isdigit(*token)) {
            Node* right = createNumber(strtol(token, NULL, 10));
            Node* current = root;
            while(current->right != NULL) {
                current = current->right;
            }
            current->right = right;
        } else {
            Node* right = root->right;
            Node* left = root->left;
            root->right = createOperator(*token, left, right);
            root = root->right;
        }
        token = strtok(NULL, " ");
    }
    return applyOp(root->value, root->right->value, root->op);
}

int main() {
    char input[50];
    printf("Enter an arithmetic expression without parentheses (e.g., 4 5 +): ");
    fgets(input, sizeof(input), stdin);
    input[strlen(input) - 1] = '\0'; // remove newline character
    int result = evaluateExpression(input);
    printf("Result: %d\n", result);
    return 0;
}